lab 2 hydraulics report

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Centennial College *

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241

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Mechanical Engineering

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Feb 20, 2024

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Course Name: Robo 241 Section: 6 Lab No.: 2 Lab report must be handed in at the beginning of your next lab class. Late submissions will result in reduction of marks. Student Name: _ ADU OLUWATENIAYO ___ (Last name) (First Name) Student No. ___301236251_____

Date: __01/23/2024____ Instructor: Instructor’s Signature: Power unit Abstract. The study of a hydraulic power unit in a lab environment explores important variables including pressure, flow rate, and efficiency. The lab obtains significant insights into the operational efficiency and performance characteristics of the unit by carefully examining these critical variables. Empirical data of this kind is fundamental to the optimization of hydraulic systems used in many industrial applications. By means of methodical experimentation and analysis, the laboratory not only augments comprehension of hydraulic principles but also furnishes significant insights that can guide the conception, functioning, and upkeep of hydraulic systems, consequently enhancing dependability, efficiency, and sustainability throughout various industrial domains.. Objectives: • Gain a comprehensive understanding of the various components within a hydraulic power unit.

• Learn the practical aspects of adjusting the swashplate angle in hydraulic pumps to optimize flow rate. • Gain practical insights into selecting appropriate pipe sizes to ensure efficient flow rate while minimizing pressure losses. Observation: 1. Calculate the theoretical flow rate of the pump at 1800 RPM. 2. 𝑄𝑡 == 16.1 𝐺𝑃𝑀 2. Calculate the heat generated if the maximum measures flow(Q) of the pump is returned to tank through the relief valve. 𝐻𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 = 16.5 × 1200 × 1.5 = 297000𝐵𝑇𝑈 3. Calculate the torque (TT) at the relief valve setting of 1200 psi. 𝑇𝑇 = = 393.4 𝑖𝑛 − 𝑙𝑏𝑠 4. Calculate the HP at maximum flow (Q) and the relief valve setting pressure (P). 𝐻𝑃 = = 11.3𝐻𝑃 5. Calculate the volume of oil in the reservoir, assuming it is 75% full. 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 = 47.25 × 27 × 15 = 19136.25 𝑖𝑛 3 = 19136.25 𝑖𝑛 3 × 0.75 = 14352.1875

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𝑉𝑜𝑙𝑢𝑚𝑒 𝑖𝑛 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 = = 62.1 𝐺𝑃𝑀 6. Velocity of oil through the pressure line and inlet pipe (assuming schedule 80) at the measured maximum pump delivery. 𝐼𝐷 = (1.050) 2 × 0.7854 = 0.865 𝑠𝑞 − 𝑖𝑛 𝑉 = 0.3208 × 16.1 × 0.865 = 4.46 𝑓𝑡 − 𝑠 7. Calculate the heat which can be radiated from the surface pf the reservoir per hour in the hydraulic fluid temperature is 115F and the ambient room temperature is 72F. 1𝐻𝑃 = 42.4 𝐵𝑇𝑈 = 11.3 × 42.2 × 60 = 28,675 𝐵𝑇𝑈 8. State the electric motor has adequate power to run the pump if the pressure is raised to 2500 psi. 𝑄 × 𝑝 𝑝𝑜𝑤𝑒𝑟(𝑘𝑤) = 1714 𝑃 = = 11𝑘𝑤

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Conclusion; • The hydraulic power unit requires 11.3 horsepower (HP) at the relief valve setting and the measured flow rate. • The reservoir needs approximately 62.1% gallons to reach 75% capacity, which is the standardized amount in the industry. • The electric motor has been calculated to have adequate power (11KW) to run the pump even if the pressure is raised 20 2500 psi. • The system generates heat, with a calculated value of 28,675 BTU per hour. It is essential to discuss whether the level of heat is desirable for the optimal operation of the hydraulic unit. Heat management is critical to prevent system overheating and ensure long-term reliability.

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• The calculated oil velocities through the pressure line and inlet pipe were found to be 4.46 feet/s. This velocity is well within the acceptable range, as it is significantly below the established maximum limit of 20 f/s. In summary, addressing these points which i fully understood provides a comprehensive overview of the hydraulic power unit’s performance, ensuring the key aspects such as power requirements, reservoir volume, oil velocities, motor power, and heat generation.